Parametric Equations: Differentiation and Integration | 参数方程：微分与积分

📚 Parametric Equations: Differentiation and Integration | 参数方程：微分与积分

hematics, you are expected to master differentiation and integration of parametric forms, including finding tangents, normals, and areas. This article covers all essential concepts, worked examples, and exam-ready tips.

参数方程提供了一种强大的方式来描述那些难以用单一函数 y = f(x) 表示的曲线。在 A-Level Edexcel 数学考试中，你需要掌握参数形式的微分与积分，包括求切线、法线以及面积。本文涵盖所有核心概念、例题解析及考试技巧。

1. What Are Parametric Equations? | 什么是参数方程？

In parametric equations, the coordinates x and y are both expressed as functions of an independent parameter, commonly t or θ. Instead of linking y directly to x, we describe how each coordinate changes with the parameter. This is particularly useful for curves that loop or have multiple y‑values for a single x.

在参数方程中，坐标 x 和 y 都表示为独立参数（通常用 t 或 θ）的函数。我们不再直接将 y 与 x 联系起来，而是描述每个坐标如何随参数变化。这对于那些存在回路或同一个 x 对应多个 y 值的曲线特别有用。

n t for 0 ≤ t < 2π. Here t represents the angle measured anticlockwise from the positive x‑ ich is often easier to handle than the Cartesian equivalent y² = 4ax when differentiating.

例如，单位圆可表示为 x = cos t, y = sin t, 0 ≤ t < 2π。这里 t 表示从正 x 轴逆时针测量的角度。同样，抛物线可写作 x = 2at, y = at²，在求导时这通常比笛卡尔形式 y² = 4ax 更容易处理。

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2. Converting to Cartesian Form | 转化为笛卡尔形式

er, or by using trigonometric identities like cos²t + sin²t = 1. The choice of method depends on the form of the parametric functions.

为得到笛卡尔方程，我们需要消去参数 t。可以通过从一个参数方程解出 t 代入另一个，或利用三角恒等式如 cos²t + sin²t = 1 来实现。具体方法取决于参数函数的形式。

tric cases such as x = 3 cos θ, y = 4 sin θ, we rewrite cos θ = x/3 and sin θ = y/4, then use cos²θ + sin²θ = 1 to obtain x²/9 + y²/16 = 1, an ellipse.

若 x = 2t + 1 且 y = 4t²，由第一式可得 t = (x − 1)/2，代入 y 得到 y = 4[(x − 1)/2]² = (x − 1)²，这是一条标准抛物线。对于三角函数情形，如 x = 3 cos θ, y = 4 sin θ，可改写为 cos θ = x/3, sin θ = y/4，然后利用 cos²θ + sin²θ = 1 得到 x²/9 + y²/16 = 1，这是一个椭圆。

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3. Sketching Parametric Curves | 绘制参数曲线

When sketching a parametric curve, it is often helpful to prepare a table of values for t and the corresponding (x, y). Plot these points and join them in increasing order of t. Always indicate the direction of increasing parameter with arrows on the curve – this is a common requirement in exam questions.

绘制参数曲线时，通常需要先制作一个关于 t 和相应 (x, y) 的数值表。描出这些点，并按照 t 增大的顺序将它们连接起来。务必在曲线上用箭头标出参数增大的方向——这是考试中常见的得分点。

ates. The curve may cross itself; this does not matter as long as you follow the parameter' the curve corresponds to which interval of t and is crucial for later integration limits.

考虑 x = t² − 1, y = t(3 − t)。选取 t = −1, 0, 1, 2, 3 得到一系列坐标。曲线可能会自交，这没关系，只需按照参数的推进绘制。方向有助于识别曲线的哪一部分对应哪个 t 区间，这对后续的积分限至关重要。

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4. Differentiation: Finding dy/dx | 微分：求 dy/dx

etric equations x = f(t), y = g(t), the derivative dy/dx is obtained using the chain rule:

dy/dx = (dy/dt) / (dx/dt)

这是参数方程 x = f(t), y = g(t) 中求 dy/dx 的公式，利用链式法则得出。

/dt = 3t² − 1, giving dy/dx = (3t² − 1)/(2t). Notice that the gradient is expressed in terms of the parameter t, not x – to find the gradient at a specific point you must first find the t‑value that produces those coordinates.

该公式适用于 dx/dt ≠ 0 的情况。先分别求出 dx/dt 和 dy/dt，然后相除。例如，若 x = t² + 1, y = t³ − t，则 dx/dt = 2t, dy/dt = 3t² − 1，所以 dy/dx = (3t² − 1)/(2t)。注意梯度是以参数 t 表示的，而非 x——若要求某一点处的梯度，必须首先找到产生该坐标的 t 值。

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5. Second Derivative d²y/dx² | 二阶导数

To find the second derivative, differentiate dy/dx with respect to x again. However, since dy/dx is a function of t, we must use the chain rule once more:

d²y/dx² = d/dx (dy/dx) = ( d/dt [dy/dx] ) / (dx/dt)

要求二阶导数，需对 dy/dx 再次关于 x 求导。但由于 dy/dx 是 t 的函数，必须再次运用链式法则：d²y/dx² = d/dx(dy/dx) = (d/dt[dy/dx]) / (dx/dt)。

4t²) = (12t² − 6t² + 2) / (4t²) = (6t² + 2)/(4t²) = (3t² + 1)/(2t²). Then divide by dx/dt = 2t to get d²y/dx² = (3t² + 1)/(2t²) ÷ (2t) = (3t² + 1)/(4t³).

继续前面的例子，dy/dx = (3t² − 1)/(2t)。对其关于 t 求导：设 u = 3t² − 1, v = 2t，则 = (6t·2t − (3t² − 1)·2)/(4t²) = (12t² − 6t² + 2)/(4t²) = (6t² + 2)/(4t²) = (3t² + 1)/(2t²)。再除以 dx/dt = 2t 得到 d²y/dx² = (3t² + 1)/(2t²) ÷ (2t) = (3t² + 1)/(4t³)。

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6. Equations of Tangents and Normals | 切线与法线方程

get the gradient m. The tangent equation is y − y₁ = m(x − x₁). The normal gradient is −1/m (provided m ≠ 0), and the normal line can be written similarly.

要写出参数曲线上某点的切线方程，首先找出该点对应的 t 值，然后计算该 t 值下的 dy/dx 得到梯度 m。切线方程为 y − y₁ = m(x − x₁)。法线梯度为 −1/m（只要 m ≠ 0），类似可写出法线方程。

cos t, y = sin 2t. Find the tangent at t = π/3. First x = 2 cos(π/3) = 1, y = sin(2π/3) = √3/2. Then dx/dt = −2 sin t, dy/dt = 2 cos 2t. At t = π/3, dx/dt = −√3, dy/dt = 2 cos(2π/3) = −1. So dy/dx = (−1)/(−√3) = 1/√3. Tangent: y − √3/2 = (1/√3)(x − 1). Normal: gradient −√3, equation y − √3/2 = −√3(x − 1).

例如，曲线由 x = 2 cos t, y = sin 2t 定义，求 t = π/3 处的切线。首先 x = 2 cos(π/3) = 1, y = sin(2π/3) = √3/2。dx/dt = −2 sin t, dy/dt = 2 cos 2t。在 t = π/3 处，dx/dt = −√3, dy/dt = 2 cos(2π/3) = −1。故 dy/dx = (−1)/(−√3) = 1/√3。切线：y − √3/2 = (1/√3)(x − 1)。法线：梯度 −√3，方程 y − √3/2 = −√3(x − 1)。

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7. Integration: Area Under a Parametric Curve | 积分：参数曲线下的面积

When finding the area bounded by a parametric curve and the x‑axis, we cannot directly integrate y with respect to x if the relationship is complicated. Instead, we replace dx by (dx/dt) dt and integrate with respect to t:

Area = ∫ y dx = ∫_t=α^t=β y (dx/dt) dt

求参数曲线与 x 轴围成的面积时，如果直接对 x 积分 y 很复杂，我们可以将 dx 替换为 (dx/dt) dt，并关于 t 积分：Area = ∫ y dx = ∫ y (dx/dt) dt。

The limits α and β are the parameter values that correspond to the x‑limits of the region. It is essential to keep the limits consistent with the orientation of the curve. If the curve is traced once as t increases from α to β, the integral gives the signed area under the curve.

积分限 α 和 β 是对应该区域 x 范围的参数值。保持积分限与曲线的方向一致至关重要。如果当 t 从 α 增大到 β 时曲线恰好描过一次，该积分给出曲线下的有号面积。

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8. Limits of Integration in Parametric Form | 参数形式中的积分限

Determining the correct t‑limits is often the trickiest part. You must solve the parametric equations to find the t‑values corresponding to the x‑intercepts or the endpoints of the region. If the curve doubles back on itself, you may need to split the integration into parts.

确定正确的 t 积分限往往是最棘手的部分。你需要解参数方程，找到与 x 截距或区域端点对应的 t 值。如果曲线有折回，可能需要将积分分段处理。

he loop occurs where the curve crosses itself; solve y = 0 for the intersection with the x‑axis: t(t² − 1) = 0 ⇒ 1. The loop is traced from t = −1 to t = 1, but notice that for t between −1 and 0, the x‑values first decrease then increase? rom t=0 to 1 of y (dx/dt) dt with appropriate sign. Always check the direction of tracing.

例如，求曲线 x = t² − 1, y = t³ − t 所围成环的面积。该环出现在曲线自交处；令 y = 0 求出与 x 轴的交点：t(t² − 1) = 0 ⇒ t = −1, 0, 1。环从 t = −1 到 t = 1 被描出，但需注意 t 在 −1 到 0 之间时 x 的变化方向。实际画图可知环是对称的，面积可计算为 2 × ∫₀¹ y (dx/dt) dt ，并注意符号。始终检查描图方向。

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9. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Many students forget to divide by dx/dt when finding dy/dx, or they mishandle the second derivative. Another frequent error is using the wrong t‑limits for area integration – always map the x‑range back to the parameter using the function x(t) and ensure the limits are in the correct order (lower limit corresponds to the smaller t if t increases with x).

许多学生在求 dy/dx 时忘记除以 dx/dt，或在处理二阶导数时出错。另一个常见错误是在面积积分时使用了错误的 t 限——务必利用 x(t) 将 x 范围映射回参数，并确保积分限顺序正确（如果 t 随 x 增大而增大，则下限对应较小的 t）。

esian form suggests a full curve, but the parametric definition only covers part of it. Finally, always show the steps of differentiation and quote the formula dy/dx = (dy/dt)/(dx/dt) to secure method marks.

在消去参数时，注意 t 的取值范围，因为它可能限制笛卡尔方程的定义域。考官常设陷阱：笛卡尔形式看起来像是完整的曲线，但参数定义只涵盖了其中一部分。最后，务必展示微分步骤并引用公式 dy/dx = (dy/dt)/(dx/dt)，以确保拿到方法分。

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10. Worked Example: Full Solution | 例题完整解析

y = 1 − cos 2t, for 0 ≤ t ≤ π. Find the area of the region bounded by the curve and the x‑axis.

一曲线由参数方程 x = 2t + sin 2t, y = 1 − cos 2t 定义，0 ≤ t ≤ π。求该曲线与 x 轴所围成区域的面积。

Solution: First, find the t‑limits where the curve meets the x‑axis. Set y = 0: 1 − cos 2t = 0 ⇒ cos 2t = 1 ⇒ with t (dx/dt = 2 + 2 cos 2t ≥ 0), the limits for t are 0 to π. The area is given by A = ∫_t=0^π y (dx/dt) dt.

解：首先，找出曲线与 x 轴相交的 t 限。令 y = 0: 1 − cos 2t = 0 ⇒ cos 2t = 1 ⇒ 2t = 2nπ。在 0 ≤ t ≤ π 内，解为 t = 0 和 t = π。当 t = 0 时 x = 0；t = π 时 x = 2π。由于 x 随 t 增大而增大（dx/dt = 2 + 2 cos 2t ≥ 0），t 的积分限为 0 到 π。面积 A = ∫₀^π y (dx/dt) dt。

integrand is y dx/dt = (2 sin²t)(4 cos²t) = 8 sin²t cos²t = 2 × (4 sin²t cos²t) = 2 sin²2t? ctually 4 sin²t cos²t = (2 sin t cos t)² = sin²2t. So 8 sin²t cos²t = 2 sin²2t. Thus A = ∫₀^π 2 sin²2t dt.

计算 dx/dt = 2 + 2 cos 2t = 2(1 + cos 2t) = 4 cos²t（利用恒等式 1 + cos 2t = 2 cos²t）。同时 y = 1 − cos 2t = 2 sin²t。因此被积函数 y dx/dt = (2 sin²t)(4 cos²t) = 8 sin²t cos²t = 2 sin²2t。故 A = ∫₀^π 2 sin²2t dt。

Use the power‑reduction formula sin²2t = (1 − cos 4t)/2. Then A = ∫₀^π 2 × (1 − cos 4t)/2 dt = ∫₀^π (1 − cos 4t) dt = [t − (sin 4t)/4]₀^π = π − 0 = π. The required area is π square units.

利用降幂公式 sin²2t = (1 − cos 4t)/2，则 A = ∫₀^π 2 × (1 − cos 4t)/2 dt = ∫₀^π (1 − cos 4t) dt = [t − (sin 4t)/4]₀^π = π − 0 = π。所求面积为 π 平方单位。

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